Question

Find the sum of all two digit odd numbers.

Answer 1

Smallest two digit odd number = 11
Largest two digit odd number = 99
Difference between consecutive odd numbers = 2

$\therefore \text{The sequence is}11,13,....,99$

The number of terms in the given AP is
$a_n=a+(n-1)d$
$\Rightarrow 99=11+(n-1)2$
$\Rightarrow (n-1)2=88$
$\Rightarrow n-1=44$
$\Rightarrow n=45$

Sum of two digit odd numbers is,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow \frac{45}{2}\left[2\right(11)+(45-1\left)2\right]$
$\Rightarrow \frac{45}{2}[22+88]$
$\Rightarrow 45\left[55\right]=2475$